Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{-5x - 20}{x^2 - 4x - 45} \times \dfrac{6x + 30}{-3x - 12} $
First factor the quadratic. $a = \dfrac{-5x - 20}{(x + 5)(x - 9)} \times \dfrac{6x + 30}{-3x - 12} $ Then factor out any other terms. $a = \dfrac{-5(x + 4)}{(x + 5)(x - 9)} \times \dfrac{6(x + 5)}{-3(x + 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ -5(x + 4) \times 6(x + 5) } { (x + 5)(x - 9) \times -3(x + 4) } $ $a = \dfrac{ -30(x + 4)(x + 5)}{ -3(x + 5)(x - 9)(x + 4)} $ Notice that $(x + 4)$ and $(x + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -30(x + 4)\cancel{(x + 5)}}{ -3\cancel{(x + 5)}(x - 9)(x + 4)} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $a = \dfrac{ -30\cancel{(x + 4)}\cancel{(x + 5)}}{ -3\cancel{(x + 5)}(x - 9)\cancel{(x + 4)}} $ We are dividing by $x + 4$ , so $x + 4 \neq 0$ Therefore, $x \neq -4$ $a = \dfrac{-30}{-3(x - 9)} $ $a = \dfrac{10}{x - 9} ; \space x \neq -5 ; \space x \neq -4 $